This is an article picked up in sci.physics which describes similarities between the behaviour of Rubics cube and the spin of quarks (!)

Crude conversion to hypertext by me (Oskar Enoksson).

>From: (Michael Weiss) >Newsgroups: sci.math >Date: 12 Apr 1995 14:46:02 GMT

Part I

Back when Rubik's cube was all the rage, Solomon Golomb noticed something remarkable: the mating habits of quarks mirror the twisting properties of the `corner cubies'. ("Rubik's Cube and a model of quark confinement", Amer. J. Phys., 49 (11), November 1981).

If you've never played with the Cube (as rubicionados call it), you should probably just go on to the next post--- I won't attempt a verbal description! But even if you have, you might not have heard of Golomb's observation. So let me explain. (You might also want have a look-see at Douglas Hofstadter's two columns on Rubik's Cube, reprinted in his book "Metamagical Themas".)

Rubik's cube is made up of 26 so-called `cubies': 8 corner cubies, 12 edge cubies, and 6 face-centered cubies. As you twist the Cube around, the face-centered cubies rotate in place, and the other cubies move from place to place while changing orientation. Instead of `place to place', the usual lingo says 'cubicle to cubicle'--- the cubicles are the little bits of space occupied by the cubies.

This is tailor-made for group theory, and we let Rubik's group be the group of all possible operations you can perform on the Cube. (Well, not *all* possible operations--- we don't include smashing it in frustration! [It's not invertible! --John Baez].) Note that an element of the Group (as I'll call it) permutes the cubies, but this permutation doesn't completely specify the group element. In fact, some group elements don't evict *any* cubies from their cozy home cubicles, but merely reorient them in place, like so many couch potatoes. The set of these group elements forms a subgroup--- I'll call it the Subgroup.

OK, let's consider an element of the Subgroup. Looking at a corner cubie, we have three possibilities: no motion at all, or a counterclockwise twist, or a clockwise twist. We now have a conservation law, of a sort. If we count clockwise twists as +1, counterclockwise as -1, and no motion as 0, then:

(I'm not saying this is obvious!) Shall we call this the Law? Perhaps not. I'll call it Rubik's Rubric. (Webster: "ru-bric authoritative rule...")

A similar law holds for the sum of the flips of the edge cubies, mod 2. But this smacks of parity, and parity is a deal less exotic than triality. So I'll just forget about edge cubies from now on. Pretend we have, not a Rubik's Cube, but a 2x2x2 Cube--- a Twobik Cube (credit Hofstadter with the pun).

And we are now ready for Golomb's observation. Mesons are made with a quark and an anti-quark, and baryons with three quarks or three anti-quarks; well, an element of the Subgroup can create a twist plus an anti-twist, or three twists or anti-twists, but never an isolated twist. (Guess we should call this the Toy Model of Quark Confinement!)

You're probably muttering, "This is sheer coincidence!" (Can one mutter an exclamation point?) Indeed, Golomb doesn't claim a deep connection--- even though group theory plays a prominent role both in Cube theory and QCD. Recently, though, I was reading Sternberg's "Group Theory and Physics", and noticed to my surprise that the theory of principal bundles (also a corner-stone of QCD) furnishes a neat proof of Rubik's Rubric. If you already know all about bundles, then I think you'll like this example; and if you don't, it's a fun place to start learning!

(Sternberg's book serves up lots of other juicy examples. If you like dodecahedra, you'll enjoy his treatment of the infrared and Raman spectroscopy of the buckyball, C_60, by way of Frobenius reciprocity. You'll find nice points of contact with John Baez's essay "Six" as well.)

Part II

OK, how do we prove Rubik's Rubric? Say g is an element of the Group, say c is a cubie, and say c stays in its own cubicle under the action of g. I'll write twist(g,c) for the element of Z_3 that says how c is twisted: 0 for no twist, +-1 for clockwise/counter-clockwise twists. So the claim is that if g is an element of the Subgroup, then An obvious line of attack: define twist(g) to be the sum on the left-hand side of this equation; then show that twist(gh) = twist(g)+twist(h) (i.e., twist is a homomorphism) and show that twist(g)=0 for some set of g's that generate the Subgroup.

The homomorphism property is pretty obvious. Not so a set of zero-total-twist generators for the Subgroup. On the other hand, there *is* a natural set of six generators for the Group: the 90 degree turns of the six faces. (Each generator moves four cubies on Twobik's Cube.)

The problem is the natural generators of the Group *do* move the cubies around. And it's not obvious how to define twist(g,c) (and hence twist(g)) when cubies move.

Bundle theory to the rescue! (If you already know about principal bundles, skip right to part II of this post.)

A *fiber bundle* has a base space, a total space, and a projection map from the total space onto the base space. There's usually more to the definition, but first let's look at the poster child of fiber bundles: the Mobius band. The total space is the set of all points on the band, and the base space is the set of all points on the meridian of the band (so the base space is a circle). To project a point on the band into the meridian, just drop a perpendicular.

The *fiber* over a base-space point is just the set of all total-space elements that project down to that point. For the Mobius band, all the fibers are line segments.

Usually, all the fibers "look the same". What does that mean? Well, it depends on what sort of bundle you're talking about (or in the usual jargon, what category you're dealing with.) More precisely, we assume that all the fibers are objects in some category, and demand that they all be isomorphic--- every category comes fully equipped out of the box with a notion of isomorphism. So you can get your daily dose of fiber from a varied menu: vector spaces, groups, topological spaces, manifolds--- personally I recommend the category-of-the-day, G-sets, with a dry white wine...

For a topological fiber bundle, one demands even more. For one thing, the total space must also be a topological space, and the projection map must be continuous. There's another condition ("local triviality") I won't bother to spell out, but for the Mobius band it amounts to this: if you snip off a little piece of the band, you get something homeomorphic to a rectangle.

Now here's a key point: while we said any two fibers must be isomorphic, we didn't say there was a "natural" or "canonical" or "preferred" isomorphism between a pair of fibers. In a way, that's the whole point of bundle theory. Take the Mobius band: if you "slide a fiber a little bit along the meridian", you get a homeomorphism between the fiber in the old position and the fiber in the new position. But if you slide the fiber smoothly once around the band, it *flips over*. Even without working through the details, it's clear we have two competing homeomorphisms of the fiber to itself: the identity map, and the flipping map.

For the Cube, discreteness reigns. So forget about topology. But we do have some groups to consider. The category of choice is now *sets with a group-action*. This calls for a new batch of definitions.

Let S be a set, and let G be a group of permutations of S. In modern jargon, one says that G acts on S. If you think of g in G as a function, you write g(s) for the result of g acting on s, but it turns out to be more convenient to write actions like multiplication. We have a choice of notation:

Say g and h are in G. If you write your action on the left, then gh means "do h, then g"; if you write it on the right, then gh means "do g, then h".

It turns out that when the action really gets hot and heavy, you want keep *both* notations handy. For example, you might have *two* groups of permutations, H and K, with H acting on S on the left and K on the right.

OK, let's say G acts on S on the right. So (sg)h = s(gh). Now there are two particularly interesting properties an action can have:

These are sort of like being onto and one-to-one. Indeed, if the action of G is both transitive and free, then we can set up a 1-1 correspondence between G and S quite easily: pick some fixed s_0 in S, and let g correspond to (s_0)g.

A *principal G-bundle* is a fiber bundle where the fibers are sets on which G acts both freely and transitively. G is called the *structure group* of the bundle.

Example? The Cube--- as we will see in

Part III

Picking up where we left off, we were about to see how the Twobik Cube is an example of a principal bundle. The structure group is Z_3, the integers mod 3.

The base space is the set of cubicles. The fiber over a cubicle is the set of three ways a cubie can sit in that cubicle. To picture these, it's convenient to label the three visible faces in some arbitrary way, say x, y, and z. (But this labelling is not itself part of the fiber!) So the three elements of the fiber might look something like this:

That dot in the middle stands for the vertex of the cubie; I haven't tried to draw the edges. So Z_3 cyclically permutes these three elements, twisting the cubie from one orientation to another. There are eight fibers (remember we have only corner cubicles), and so the total space of the bundle consists of the 8*3=24 possible positions for a cubie.

You can see how any two fibers are isomorphic. Pick a random element of the first fiber; call it p_1. Pick a random element of the second fiber; call it q_1. Now give p_1 and q_1 each a clockwise twist, and call the results p_2 and q_2. One more twist gives us p_3 and q_3, and the isomorphism is

You can construct three different isomorphisms this way. Nothing singles out any one of these isomorphisms as the preferred one.

So what does this have to do with Rubik's Rubric? Remember the problem we faced: if g is an element of the Group (*not* of Z_3), and c is a cubie, we wanted to define twist(g,c), the amount that cubie c is twisted by g.

Our cubie has become an element of the fiber over its cubicle. How about g? Elements of the Group move cubies from one position to another. So if g is an element of the Group, then g defines a map from the total space to itself.

The Group is, in fact, a subgroup of the *automorphism group* of the bundle. This just means that if you take a cubie, and twist it and then apply a Group element, you get the same result as when you apply the Group element first and the twist second. A little notation helps: let t be in Z_3, let g be in the Group, and let c be a cubie position (an element of the total space). We let Z_3 act on the right and the Group on the left, so ct stands for c with a twist, and gc stands for c after being moved by g. So the automorphism property takes the pretty form:

Now how about defining twist(g,c)? If gc happens to belong to same fiber as c, then no problem: g just twists c by some amount. In other words, gc = ct for a unique t in Z_3, and we can define twist(g,c)=t.

But the hard part remains: what if gc is in a different fiber? In this case, we need some reference cubie in the new fiber to compare gc to. It seems we need a "global coordinate system" for the bundle.

Bundle theory interrupts: "Don't use a coordinate system, use a *section* of the bundle!" (Such has been one of the lessons of bundle theory over the years: coordinate systems are more-or-less equivalent to sections. Which you use is often a matter of taste, but modern mathematical fashion much prefers to talk of sections.)

A *section* of a bundle simply picks an element from each fiber. Some more notation: if s is a section and x is a point in the base space, then s_x is the chosen element of the fiber over x. So a section in our example assigns a cubie position to each cubicle.

The Group is pretty active: it acts on cubicles, on cubie positions, and on sections. I won't write out descriptions, just a little notation. Let g be a Group element, x a cubicle, s a section, and s_x (as we just said) a cubie position in the fiber over x. Then:

That last equation may look like a jumble of symbols the first time you see it, but after working through it, I'll bet you say "Of course!"

Still sticking with the same meanings for g, s, and x, define:

or more formally: Hmm, those parameters to the twist function keep piling up... Well, let's check out the dependence on the choice of section s. Let's say r and s are sections. You can turn r into s by twisting each cubie in place. In other words, we have yet another group to worry about: the direct product of eight copies of Z_3, one for each cubicle. Letting this new group (call it T) act on the right, we have rt = s, for some t in T. Borrowing some notation, we let t_x stand for the twist at cubicle x. (Notice that the Subgroup is a subgroup of T. Note also that twist(g,s,x), as x varies over the eight cubicles, defines an element of T.)

OK, let's compare twist(g,r,x) with twist(g,s,x). We need to keep track of old occupants and new occupants. Let's say the new occupant at cubicle x originally resided at cubicle y. (Really you should work this out for yourself. But I'm feeling generous today, so here goes.) A diagram may help:

E.g., with the r section, r_x was the old occupant at x, and g(r_y) is the new occupant.

We want to compute the twist between the old and new occupants. The following diagram says it all:

Single-weight arrows (--->) stay in the same fiber, double-weight arrows (===>) move from fiber y to fiber x. The two vertical arrows on the right are both labelled t_y because g[(r_y)(t_y)] = [g(r_y)](t_y) --- g is a bundle automorphism, remember! Look now at two ways to get from s_x to g(s_y): take the twist(g,s,x) route, or take t_x in reverse, then twist(g,r,x), then t_y down. So we have the following equation in Z_3: So twist(g,s,x) is *not* independent of s. But here's the punchline: summing over x removes the dependence. For in the sum, we add t_x and subtract t_x once for each x.

And that's all we need! We *don't* ever define twist(g,x)--- we go directly to twist(g), our goal all along.

The rest is a smooth coast down the bunny slope. If you pick the right section, a generator g of the Group (a 90 degree twist of one face) has twist(g,s,x) = 0 for all x. As for the homomorphism property:

the trick here is to pick one section s to compute twist(h) and twist(gh), and another section to compute twist(g). Work it out if you're skeptical. QED (or `\omicron\epsilon\delta, as Euclid wrote).

Part IV (Last part>

Three groups played a role in our little drama: Z_3, the Group, and a group I called T earlier--- the direct product of eight Z_3's.

Z_3 is called the structure group of the bundle. We saw that the Group was a subgroup of the automorphism group of the bundle, Aut(E). The group T, which twists cubies in place, also has a special name: it's called the *gauge group* of the bundle. So the Subgroup is a subgroup of the gauge group.

Now, why did things work out so nicely? A mathematician's question. A story has it that a grad student was walking on the campus, when he met a jubilant physics prof. Said the prof, "I finally finished my calculations, and everything magically clicked into place! Now I understand it all!" The student congratulated him, and walked on. Next he met a gloomy math prof, who said, "I finally finished my calculations, and everything magically clicked into place! I don't understand it at all!"

In this case, the answer is: because the structure group is abelian. That's really all we need. If you trace through the argument carefully, you'll find it depends on three things:

If the base space is *not* finite, I imagine you can push the argument through with an integral instead of a sum--- provided you pick the right categories, of course!

Now the structure group of QED is U(1), an abelian group. Perhaps Rubik's Rubric corresponds to some well-known fact of E&M, maybe conservation of charge?

But quarks--- well, the big thing about quarks is that the structure group, SU(3), is *not* abelian. (If there's anything big about quarks!)

So my feeling was that Golomb was quarking up the wrong tree. And the skeptical reader, who doubted that Rubik's Cube could have anything to do with quarks, seems to have the last laugh.

Except then I got these tantalizing messages from John Baez:

> One other little observation... Z_3 is deeply related to > SU(3), because the center of SU(3), consisting of the diagonal > matrices exp(2 pi i n/3), is isomorphic to Z_3. More generally, > the center of SU(n) is Z_n in the same way. For SU(2) it's > Z_2, and when we mod out by this Z_2 we get good old SO(3).

> From this one would suspect that Z_3 representation theory is > important in SU(3) representation theory. My guess: Each irrep of > SU(3) is obviously (or should I say Schurly) an irrep of Z_3, > of which there are 3, and to tell which one, just count > the number of boxes in your Young diagram mod 3. I'm finally > really learning about Young diagrams and I love 'em!

> The corresponding Z_2 invariant of SU(2) irreps is just > the "integer/half-integer spin" distinction!

> You could add, if you wished, that I later learned > that 't Hooft's argument for confinement indeed relies heavily > on Z_3 as the center of SU(3), and that the details are in > G. 't Hooft, Nucl. Phys. B128 (1978) 1.

> In some sense the recent Seiberg-Witten work may eventually > lead to a clarification of 't Hooft's old ideas, and > Martellini here has detailed schemes as to how to get > it all to work... but I will try to explain all this in > a forthcoming Week.

Author: (Michael Weiss)

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